Find one value of $x$ that is a solution to the equation: $(x^2-7)^2+2x^2-14=0$ $x=$
We could solve for $x$ by expanding $(x^2-7)^2$, combining terms that are alike, and using the quadratic formula or factoring to solve for $x$. However there is a shorter and more elegant way to approach this problem. Let's use structural features to rewrite the equation in a simpler form. Note that $2x^2-14=2({x^2-7})$. This means that we can rewrite the equation as: $({x^2-7})^2+2({x^2-7})=0$ If we let ${p}={x^2-7}$, we can see that this equation is in the form: ${p}^2+2{p}=0$ Let's solve this equation in terms of ${p}$ : $\begin{aligned}{p}^2+2{p}&=0\\\\ {p}({p}+2)&=0\\\\ {p}=0\ &\text{or} \ \ {p}=-2 \end{aligned}$ Since ${p}={x^2-7}$, let's substitute this value back into our two solutions in order to solve for $x$ : ${x^2-7}=0\ \ \ \text{or} \ \ \ {x^2-7}=-2$ When we solve $x^2-7=0$, we find that $x=\pm\sqrt{7}$. When we solve $x^2-7=-2$, we find that $x=\pm\sqrt{5}$. In conclusion, the four solutions of the equation $(x^2-7)^2+2x^2-14=0$ are: $x=\sqrt{7}$ $x=-\sqrt{7}$ $x=\sqrt{5}$ $x=-\sqrt{5}$ [Is there another way to solve for x?]